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The length of a spring is increased by 2.5 cm the elastic potential energy stored in the spring is U . If the length is increasedby 5 cmits elastic potential energy become …… |
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Answer» SOLUTION :The elastic potential energy STORED in the springis `U = 1/2 ky^(2)` ` :. U PROP y^(2)` ` :. (U_(2))/(U_(1)) = ((y_(2))/(y_(1)))^(2) = ((5.0)/(2.5))^(2) = (2)^(2) = 4 ` ` U_(2) = 4U "" [ :. U_(1) = U] ` |
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